What is the integral of f(x)=(x-6)(x+9)(x-8) ?

The integral of f(x)=(x-6)(x+9)(x-8) is (x^4)/3-(x^4}{12}-4x^3+\frac{7x^3)/3-39x^2+432x+C

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integral of f(x)=(x-6)(x+9)(x-8)

integral

f (x) = (x - 6) (x + 9) (x - 8)

\frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{7 x ^{3}}{3} - 39 x^{2} + 432 x + C

Solution steps

\int (x - 6) (x + 9) (x - 8) d x

Apply Integration By Parts

: (x + 9) (\frac{x ^{3}}{3} - 7 x^{2} + 48 x) - \int \frac{x ^{3}}{3} - 7 x^{2} + 48 x d x

= (x + 9) (\frac{x ^{3}}{3} - 7 x^{2} + 48 x) - \int \frac{x ^{3}}{3} - 7 x^{2} + 48 x d x

\int \frac{x ^{3}}{3} - 7 x^{2} + 48 x d x = \frac{x ^{4}}{12} - \frac{7 x ^{3}}{3} + 24 x^{2}

= (x + 9) (\frac{x ^{3}}{3} - 7 x^{2} + 48 x) - (\frac{x ^{4}}{12} - \frac{7 x ^{3}}{3} + 24 x^{2})

Simplify

(x + 9) (\frac{x ^{3}}{3} - 7 x^{2} + 48 x) - (\frac{x ^{4}}{12} - \frac{7 x ^{3}}{3} + 24 x^{2}) : \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{7 x ^{3}}{3} - 39 x^{2} + 432 x

= \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{7 x ^{3}}{3} - 39 x^{2} + 432 x

Add a constant to the solution

= \frac{x ^{4}}{3} - \frac{x ^{4}}{12} - 4 x^{3} + \frac{7 x ^{3}}{3} - 39 x^{2} + 432 x + C

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What is the integral of f(x)=(x-6)(x+9)(x-8) ?
The integral of f(x)=(x-6)(x+9)(x-8) is (x^4)/3-(x^4}{12}-4x^3+\frac{7x^3)/3-39x^2+432x+C